91影视

A particle is executing a simple harmonic motion.

a continuous random variable,whose integral across an interval offers the likelihood that the variable's value falls inside the same interval.

Function x is given as .

$\begin{array}{c}v=\frac{dx}{dt}\\ =鈭�a蝇\sin \left(蝇t\right)\\ =鈭�a蝇\sqrt{1鈭�{\cos }^2\left(蝇t\right)}\end{array}$

The value of $\frac{dx}{dt}$is given below.

$\begin{array}{c}\frac{dx}{dt}=鈭�a蝇\sqrt{1鈭�{\left(\frac{x}{a}\right)}^2}\\ =鈭�蝇\sqrt{a^2鈭�x^2}\end{array}$

$\begin{array}{c}dt=\frac{dx}{v}\\ =鈭�\frac{dx}{蝇\sqrt{a^2鈭�x^2}}\end{array}$

The probability of function is given below.

$\begin{array}{c}f\left(x\right)dx=\frac{dx}{\sqrt{a^2鈭�x^2}}\\ f\left(x\right)=\frac{C}{\sqrt{a^2鈭�x^2}}\end{array}$

Find the value of constant.

$\begin{array}{c}{鈭�}_{鈭�a}^a\frac{Cdx}{\sqrt{a^2鈭�x^2}}=C{\left[\arccos \left(\frac{x}{a}\right)\right]}_{鈭�a}^a\\ =C蟺\\ C蟺=1\\ C=\frac{1}{蟺}\end{array}$

Substitute the value of C the function becomes as follows.

$f\left(x\right)=\frac{1}{蟺\sqrt{a^2鈭�x^2}}$

The graphs are shown below.

The average of the function is mentioned below

$\begin{array}{c}渭={鈭�}_{鈭�\mathrm{鈭�}}^{\mathrm{鈭�}}xf\left(x\right)dx\\ ={鈭�}_{鈭�\mathrm{鈭�}}^{\mathrm{鈭�}}\frac{xdx}{蟺\sqrt{a^2鈭�x^2}}\\ ={\left[鈭�\frac{\sqrt{a^2鈭�x^2}}{蟺}\right]}_{鈭�\mathrm{鈭�}}^{\mathrm{鈭�}}\\ =0\end{array}$

The variance is given below.

$\begin{array}{c}\mathrm{Var}\left(x\right)={鈭�}_{鈭�\mathrm{鈭�}}^{\mathrm{鈭�}}{x}^{2}f\left(x\right)dx\\ ={鈭�}_{鈭�a}^a\frac{x^{2}dx}{蟺\sqrt{a^2鈭�x^2}}\\ ={\left[\frac{a^2\arcsin \left(\frac{x}{\left|a\right|}\right)鈭�x\sqrt{a^2鈭�x^2}}{2蟺}\right]}_{鈭�a}^a\\ =\frac{a^2}{2}\end{array}$

The standard deviation is given below.

$\begin{array}{c}蟽=\sqrt{\mathrm{var}\left(x\right)}\\ =\sqrt{\frac{a^2}{2}}\\ =\frac{a}{\sqrt{2}}\end{array}$

The required values are mentioned below.

$\begin{array}{c}f\left(x\right)=\frac{1}{蟺\sqrt{a^2鈭�x^2}}\\ 渭=0\\ \mathrm{Var}\left(\mathrm{x}\right){}_{-\mathrm{a}}^{\mathrm{a}}=\frac{a^2}{2}\\ 蟽=\frac{a}{\sqrt{2}}\end{array}$